The Fluvial Geomorphic System Definition Variables of Stream Flow Hydrologic cycle Discharge

The Fluvial Geomorphic System• Definition

• Variables of Stream Flow

• Hydrologic cycle• Discharge• Floods• Effect of Slope, Hydraulic Radius

• Equilibrium in Streams

• Graded Stream• Degradation• Aggradation

The Fluvial Geomorphic System

How is sediment transported and removed from continents?(i.e., what mechanisms are most important

in shaping landscapes?)

► Rivers: 85-90%

► Glaciers: 7%

► Groundwater & Waves: 1-2%

► Wind: < 1%

► Volcanoes: < 1%

The fluvial system encompasses:

► Drainage divides,

► Source areas of water and sediment,

► Channels and valleys of the drainage basin,

► Depositional Areas

Example watershed--sketch

Example watershed—on shaded relief map

Example watershed—two-dimensional

Hydrologic cycle

Water budget/balance:

Inputs – Outputs = +/- Storage

Inputs? precipitation

Outputs?

runoff

evapotranspiration

GW discharge

Storage?

Soil moistureFloodingaquifer storage

Inputs – Outputs = +/- Storage

PCIP - (ET + RO + GW) = ΔS

PCIP - ET - RO - GW = ΔS

PCIP = ET + RO + GW + ΔS100% 25-40%

Hydrologic cycle

Interception = INT = ET + Evaporation + Infiltration

PCIP = RO + INT + ΔS

100% 25-40% 60-75% 0%

w

Cross-sectional area and wetted perimeter

d

Area = w x d

Wetted perimeter = w + 2d

Discharge

Cross-sectional area and wetted perimeter

w

d

Area = 2w x 2d = 4wd

Wetted perimeter = 2w + 2(2d) = 2w + 4d

2w

2d

Discharge

Area A = wd

Area B = 2w x 2d = 4wd

Area B / Area A = 4wd / wd = 4 -----------------------------------------------------------

Wetted perimeter A = w + 2d

Wetted perimeter B = 2w + 2(2d) = 2w + 4dWetted perimeter B = 2(w + 2d)

Wetted perimeter B / Wetted perimeter A =

2(w + 2d) / (w + 2d) = 2

Discharge

Cross-sectional area and wetted perimeter

w

d

• Small increase in wetted perimeter (relative to increase in area) means less frictional resistance, water can flow faster (increased velocity)

Discharge

Cross-sectional area and wetted perimeter

Result: increased discharge (Q) is caused by increases in width, depth and velocity

Q = w x d x v

W V

D

Discharge

Q = aQb x cQf x kQm

a x c x k = 1

b + f + m =1

Discharge

Floods James River in Richmond, Virginia at flood stage, November 1985. Photo by Rick Berquist, used with permission.

Time

Riv

er E

leva

tio

n

Start of rainstorm End of rainstorm

Floods

Hydrograph: a plot of river level (or discharge) versus time

Note equivalence of river elevation (stage) and discharge

Time

Riv

er E

leva

tio

n

Floods

Different watersheds display different hydrograph characteristics

larger river in large watershed

small stream

Time

Riv

er E

leva

tio

n

Prior to urbanization

Start of rainstorm End of rainstorm

Precipitation

Ground

Runoff

Infiltration

Time

Riv

er E

leva

tio

n

Prior to urbanization

Start of rainstorm End of rainstorm

Precipitation

Ground

Runoff

Infiltration

Time

Riv

er E

leva

tio

n

prior to urbanization

after urbanization

Start of rainstorm End of rainstorm

Precipitation

ImperviousGround

Increased Runoff

Little Infiltration

1993 Mississippi River Flood (500-year flood)

1993 Mississippi River Flood (500-year flood)

http://www.cgrer.uiowa.edu/research/exhibit_gallery/great_floods/wetness.html

June 6, 1993

July 15, 1993

Soil Moisture

(brighter = wetter)

July 29, 1993

1993 Mississippi River Flood (500-year flood)

Time

Riv

er E

leva

tio

n

dry soils

saturated soils

Start of rainstorm End of rainstorm

Precipitation

ImperviousGround

Increased Runoff

Little Infiltration

Constructing a rating curveFloods

Note equivalence of stage and discharge

Chester IL Rating Curve( based on annual peak flows )

1.0E+04

1.0E+05

1.0E+06

1.0E+07

0 10 20 30 40 50 60

Stage (ft)

Q (

cfs)

Example rating curve

Note that rating curve allows estimation of discharge for extreme floods.

Estimating stagelevel of past floods—can then use rating curve to estimate discharge

WATTS HOME

TWELVEPOLE CREEK

OLD CULVERT

RR TRACKS

Normal water level

STEEP VALLEY WALL

FLOOD PLAIN

~17 ft ~10-12 ft

NOT TO SCALE

~5-7 ft

BASE OF DITCH

Wayne Co. flood case

WATER LEVEL, 11/12/03

~6-8 ft

Flood Intensity

0.01

0.10

1.00

10.00

100.00

1.E+00 1.E+01 1.E+02 1.E+03 1.E+04 1.E+05 1.E+06 1.E+07

Area

Q /

Are

a

FloodsRecurrence interval (RI) is the average number of years between a flood of a given magnitude.

• For example: the 100-year flood is the stage or discharge that occurs on average every 100 years.

• Different for every river.

• Data less reliable for larger RI. Why?

• RI = (N +1) / m

• N = # of years of record , m = rank

• Example: If records were kept for 59 years (N=59), and a stage level of 52 ft was the third highest level (m=3) reached during this period, then a flood of this magnitude would be categorized as a 20-year flood (RI = 60/3).

Chester IL -- Rank versus Year

0

10

20

30

40

50

60

70

80

1930 1940 1950 1960 1970 1980 1990 2000 2010

Year

Ran

k

Example of data used to calculate RI

Miss. River, Chester, Il – 1993

Note that the probability of a flood of a given magnitude is 1/RI.

Example: In any year, the chance of a100-year flood is 1/100 = 1%

The mean annual flood is the average of the maximum annual floods over a period of years.

RImean = 2.33

Chester IL Rating Curve( based on annual peak flows )

1.0E+04

1.0E+05

1.0E+06

1.0E+07

0 10 20 30 40 50 60

Stage (ft)

Q (

cfs)

Floods James River in Richmond, Virginia at flood stage, November 1985. Photo by Rick Berquist, used with permission.

James River, Richmond VAThree largest floods recorded from 1935 to present.

1. June 23, 1972, 28.62 ft (gage height), 313,000 cfs (discharge)

2.August 21, 1969, 24.95 ft (gage height), 222,000 cfs (discharge)

3.November 7, 1985, 24.77 ft (gage height), 218,000 cfs (discharge)

•From the picture of the river at normal flow, estimate the stage at these conditions.

•Calculate RI and probability for each of these flood events

Flood Exercise

Floods Paleofloods

• Causes: dam outbursts, glacial outbursts, extreme precipitation events.• ice dam collapsed during last Ice Age in eastern Washington, emptying

lake about half size of Lake Michigan; floodwaters had Q~752,000,000 cfs.

• Provide direct evidence of extreme hydrologic events that may shed lightback to mid-Holocene (~5,000 years)

• Flood deposits and flood erosional effects are primary sources of information about the magnitude and frequency of these extreme events.

Floods Paleofloods

Example use of paleoflood records to discern mid-Holocene climates

Hirschboeck, 2003

Floods Paleoflood

“reconstruction”

• What is needed to estimate discharge, Q, during a modern flood?

•Rating curve allows Q to be estimated from stage

• What is needed to estimate discharge during a paleoflood?

• flood stage may not be known• If flood stage is known, no rating curve for extreme stage• velocity must be estimated and ancient valley shape must be estimated

Q = w x d x v

Floods Paleoflood

“reconstruction”

Methods for estimating stage of paleofloods• depositional: slack-water deposits in tributary valleys, caves, etc.

• slack-water deposits formed during sudden velocity decreases following peak discharge

• only preserved in protected areas above elevation of modern floods (“non-exceedance level”

• erosional: terrace benches, markings on paleosols, bedrock walls, etc.

• vegetation: damaged trees, etc.

Hirschboeck, 2003

FloodsPaleofloods

Floods Paleoflood

“reconstruction”

Methods for estimating velocity of paleofloods

•quantitative empirical or theoretical relationships• Chezy formula: uses hydraulic radius and slope to estimate velocity• Use sizes of boulders transported in flood to estimate velocity• Manning equation: uses hydraulic radius and slope to estimate velocity:

v = 1.49/n x R2/3 x S1/2

n = roughness factorR = hydraulic radiusS = slope Wetted perimeter (WP) = 2d + w

Area (A) = wd

R = A / WP = wd / (2d + w)

dd

w

Relationships among channel shape,velocity, slope and erosional energy

Manning equation: relates hydraulic radius and slope to velocity

v = 1.49/n x R2/3 x S1/2

n = roughness factorR = hydraulic radiusS = slope

Wetted perimeter (WP) = 2d + w

Area (A) = wd

R = A / WP = wd / (2d + w)

dd

w

Relationships among channel shape,velocity and erosional energy

Wetted perimeter (WP) = 2d + w

Area (A) = wd

R = A / WP = wd / (2d + w)

2

10

1

20

WP = 14A = 20R = 1.4

WP = 22A = 20R = 0.9

Relationships among channel shape,velocity and erosional energy

2

10

1

20

WP = 14A = 20R = 1.4

WP = 22A = 20R = 0.9

What does Manning equation say about flow in these two different channel shapes if slope and roughness are equal?

v = 1.49/n x R2/3 x S1/2

n = roughness factorR = hydraulic radiusS = slope

•larger radius means greater velocity.

• smaller radius means less velocity.

• Tendency of smaller radius to restrict velocity is result of turbulence and friction as water contacts the channel margins. This causes erosion !

Relationships among channel shape,velocity, slope and erosional energy

2

10

1

20

WP = 14A = 20R = 1.4

WP = 22A = 20R = 0.9

Hydraulic shear

• low hydraulic radius• high friction/turbulence• high scour• coarse bedload

• high hydraulic radius• low friction/turbulence• low scour• fine bedload

Relationships among slope, velocity and erosional energy

•Increased discharge causes increase in depth, width and velocity--causes moderate increase in erosion.

•Scenario might occur as a result of climate change

•Increased slope at constant discharge means velocity increases, but depth decreases—causes more dramatic increase in erosion.

• Scenario might occur as a result of uplift

Sediment load:mass of sediment transported in a stream or river per unit time

example: pounds per year

Related concepts:

• denudation rates (example: ft/1000 yrs)

• sediment yield = sediment load / area

Controls on sediment load

•topographic relief•geology of watershed•climate•vegetation•other processes in watershed (glaciers,mass wasting, etc.)

Sediment load depends on: • relief – denudation rates

increase exponentially with relief of watershed.

• vegetation – sediment yield is at maximum for about 10 in/yr of precipitation. Why?

Total sediment load =

dissolved load (50%?) + flotation + suspended + bed load

-------------------------------------------suspended load: particles supported by water column

bedload: particles suspended by channel bed

Mississippi River sediment

As discharge increases, suspended load increases at

more rapid rate than discharge.

Channel patterns

• straight (rare)

• meandering (most common)

• single channel• sinuous (Ls / Lv)• few islands• deep, narrow channels• meander size proportional to Q,

maybe load

• braided

• low sinuosity• multiple, shifting channels• islands• wide, shallow channels

Causes of meandering• laminar flow tends not to be maintained, so water is deflected, energy is distributed unequally in channel.

•cut banks•point bars

• positive feedback system

• more meandering results in wider valleys, bigger floodplains.

Braided Streams• temporary, shifting channels have prompted conclusion

that braided streams are overloaded with sediment and, in response, are aggrading.

• In fact, braiding is related to erodabilty of bank material—braiding seems to develop in easily- erodable (non-cohesive) sediments (i.e., sand & gravel). See Figures 5-36 & 5-38

• Higher silt/clay ratios of load mean lower W/D ratios, development of helical flow, resistance of banks to erosion, and meandering channel patterns.

• Lower silt/clay ratios of load mean higher W/D ratios, absence of helical flow, erosion of banks, and braided channel patterns.

• Change in silt/clay to sand/gravel bank materials may result in a change in channel shape from meandering to braided-- will mean an increase in slope.

– Why? – But change in slope is a response to change in channel shape,

not a cause of braiding!